Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x+6y &= 1 \\ 8x+9y &= 7\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $8x = -9y+7$ Divide both sides by $8$ to isolate $x$ $x = {-\dfrac{9}{8}y + \dfrac{7}{8}}$ Substitute this expression for $x$ in the first equation. $9({-\dfrac{9}{8}y + \dfrac{7}{8}}) + 6y = 1$ $-\dfrac{81}{8}y + \dfrac{63}{8} + 6y = 1$ Simplify by combining terms, then solve for $y$ $-\dfrac{33}{8}y + \dfrac{63}{8} = 1$ $-\dfrac{33}{8}y = -\dfrac{55}{8}$ $y = \dfrac{5}{3}$ Substitute $\dfrac{5}{3}$ for $y$ in the top equation. $9x+6( \dfrac{5}{3}) = 1$ $9x+10 = 1$ $9x = -9$ $x = -1$ The solution is $\enspace x = -1, \enspace y = \dfrac{5}{3}$.